EXPLANATION OF TECHNETIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 2, 2015 Technetium is a chemical element with symbol Tc and atomic number 43. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Technetium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d55s2 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of technetium (from (E1 to E3 ) are the following: E1 = 7.28 , E2 = 15.26, and E3 = 29.54 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. ' ' EXPLANATION OF Ε1 = 7.28 eV ' '= -E(5s2) + E(5s1) Here E(5s2) represents the binding energy of 5s2, while the E(5s1) represents the binding energy of 5s1. The charges (-41e) of (1s22s22p63s23p63d10 4s2 4p64d5 ) screen the nuclear charge (+43e) and for a perfect screening we would have ζ = 2. However the electrons of 5s penetrate the 4d5 and lead to the deformations of electron clouds. Thus ζ > 2. Note that 5s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since 5s1 consists of one electron, we apply the Bohr formula to write E(5s1) = (-13.6057)ζ2/n2 Therefore E1 = 7.28 eV = -E(5s2) + E(5s1) = (13.6057)ζ2 - (16.95)ζ + 4.1) / n2 Since n = 5 the above equation can be written as 0.544 ζ2 - 0.678 ζ - 7.116 = 0 Then solving for ζ we get ζ = 4.29 > 2 . Note that the two electrons of opposite spin (say the 5s2 ) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. This situation indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. ' ' EXPLANATION OF Ε2 = 15.26 eV -E(5s1) Here the E(5s1) represents the binding energy of the one electron (5s1) given by applying the Bohr formula. Therefore we write Ε2 = 15.26 eV -E(5s1) = (13.6057)ζ2 / n2 Since n = 5 we get ζ = 5.3 >2 Here 5.3 > 4.29 > 2 means that the one electron of 5s1 breaks more the symmetry of spherical shells and leads to the greater deformation of electron clouds. 'EXPLANATION OF E3 = 29.54 eV ' Here, in the absence of the one electron of 4d5 with parallel spin one must apply the Bohr formula as E3 = 29.54 eV = (13.6057) ζ2 /n2 The charges (-36e) of the electrons (1s22s22p63s23p63d104s2 4p6 ) screen the nuclear charge (+43e) and for a perfect screening we would have ζ = 7. Surprisingly using n = 4 we get ζ = 5.89 < 7 which cannot exist. In fact, the electrons of 4d5 with parallel spin under electric and magnetic mutual repulsions repel the electrons of 4p6 from symmetrical positions and lead to a perfect screening with ζ = 7. Thus using ζ = 7 we respect to find that n > 4. Under this condition we write E3 = 29.54 eV = (13.6057)72 /n2 Then solving for n we get n = 4.75 > 4. Since 4.75 < 5 we observe in the electron configuration of technetium that the two outermost electrons belong always to the 5s2 . Category:Fundamental physics concepts